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3 years ago

Which feature do all adult echinoderms have?

2 answers:

4 0

The common name to the family member of phylum Echinodermata of marine family is echinoderm. They are usually characterized by a five-fold symmetry, and possess an internal skeleton of calcite plates. They are found at every ocean depth.

The features of all adult echinoderm are:

- They have a five-fold symmetry.

- Body without segmentation.

- Spiny skin.

- Internal skeleton.

- found at every ocean depth.

BigorU [14]3 years ago

3 0

Mhmh they have smellytypes on their bodies so yeah

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Which equation is balanced?

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Answer:

2Na+F2 yields 2NaF is balanced.

Explanation:

There are 2 sodium and 2 fluorine in both reactants and product: In 2NaF the 2 is distributed because it is in the beginning of the compound.

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A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree

lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = $T_1=33.9^oC=33.9+273K=306.9 K$

Final temperature of the water = $T_2$

Specific heat capacity of water under these conditions = c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

$Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)$

$-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)$

$T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC$

The new temperature of the water bath 32.0°C.

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3 years ago

A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO

Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL$

Putting values in above equation, we get:

$2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL$

$M_2=0.336M$

Thus, the concentration of NaOH is, 0.336 M

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